Linear Transformation from R3 to R3

jolly_math
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Homework Statement
Describe explicitly a linear transformation from R3 into R3 which has as its
range the subspace spanned by (1, 0, -1) and (1, 2, 2).
Relevant Equations
linear transformation
"There is a linear transformation T from R3 to R3 such that T (1, 0, 0) = (1,0,−1), T(0,1,0) = (1,0,−1) and T(0,0,1) = (1,2,2)" - why is this the case?

Thank you.
 
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Does it help to glance at the following matrices:
$$
\begin{bmatrix}
1 & 1 & 1 \\
0 & 0 & 2 \\
-1 & -1 & 2\\
\end{bmatrix}
\times
\begin{bmatrix}
x \\
y \\
z\\
\end{bmatrix}
$$
?
 
jolly_math said:
Homework Statement:: Describe explicitly a linear transformation from R3 into R3 which has as its
range the subspace spanned by (1, 0, -1) and (1, 2, 2).
Relevant Equations:: linear transformation

"There is a linear transformation T from R3 to R3 such that T (1, 0, 0) = (1,0,−1), T(0,1,0) = (1,0,−1) and T(0,0,1) = (1,2,2)" - why is this the case?

Thank you.
A linear transformation can be fully described by its action on any basis. Can you see why?
 
Hall said:
$$
\begin{bmatrix}
1 & 1 & 1 \\
0 & 0 & 2 \\
-1 & -1 & 2\\
\end{bmatrix}
\times
\begin{bmatrix}
x \\
y \\
z\\
\end{bmatrix}
$$

For R3, would
$$
\begin{bmatrix}
1 & 1 & 1 \\
0 & 2 & 0 \\
-1 & 2 & -1\\
\end{bmatrix}
\times
\begin{bmatrix}
x \\
y \\
z\\
\end{bmatrix}
$$
also work (switching which vector corresponds to each basis)? Thanks.
 
jolly_math said:
For R3, would
$$
\begin{bmatrix}
1 & 1 & 1 \\
0 & 2 & 0 \\
-1 & 2 & -1\\
\end{bmatrix}
\times
\begin{bmatrix}
x \\
y \\
z\\
\end{bmatrix}
$$
also work (switching which vector corresponds to each basis)? Thanks.
Yes, I think. I think even double columns of ##(1,2,2)## will also satisfy the given things.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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